Campervan 12 V Circuit: Fuse, Battery Budget and a 300 W Kettle
A family is wiring the 12 V leisure circuit in a converted van. The battery is a 100 Ah lead-acid type, which should not be discharged below about 50% of its rated capacity or it will be damaged. The circuit supplies, all in parallel across the battery: a 60 W compressor fridge, two 5 W LED ceiling lights, a 36 W water pump and a 10 W phone charger. The whole positive feed passes through a single 15 A blade fuse, and the cable itself is rated for 20 A continuous. The family plans a two-night trip with no driving between the nights and no solar panel. They expect to use all the loads for about 5 hours each evening, with the fridge cycling so its compressor runs about half the time, the pump running for 12 minutes a day and the charger for 2 hours. They also want to plug a 300 W inverter into the same circuit to run a 230 V kettle, arguing that the fuse has plenty of spare capacity.
1. Calculate the total current drawn when every 12 V load is running at once, and decide whether the 15 A fuse is a sensible choice. Explain what the fuse is actually protecting.
Read the worked answer
Because the loads are in parallel, each sees the full 12 V and the currents add. Add the powers first:
P = 60 + 5 + 5 + 36 + 10 = 116 W
I = P / V = 116 / 12 = 9.7 A
This is below the 15 A fuse rating and below the 20 A cable rating, so the fuse will not blow in normal running and the cable will not overheat.
Is 15 A sensible? Yes, with one caveat. A fuse protects the cable, not the appliance: its job is to melt and break the circuit before a damaged or shorted cable overheats and starts a fire. So the fuse must sit above the largest normal current and below the cable rating, i.e. 9.7 A < 15 A < 20 A, which this does. The caveat is start-up surge: the fridge compressor and the pump motor briefly draw several times their running current, but a blade fuse has enough thermal lag to ride out a fraction of a second, so 15 A gives useful headroom without leaving the cable unprotected. A 10 A fuse would be too close to 9.7 A and could nuisance-blow on start-up.
2. Estimate whether the 100 Ah battery can support two nights of the listed use. Support your answer with an energy budget and comment on the reliability of the estimate.
Read the worked answer
Energy per evening:
- Fridge: 60 W x 5 h x 0.5 duty = 150 Wh
- Lights: 2 x 5 W = 10 W x 5 h = 50 Wh
- Pump: 36 W x 0.2 h (12 min) = 7.2 Wh
- Charger: 10 W x 2 h = 20 Wh
Total per evening = about 227 Wh, so two nights is about 455 Wh.
Usable battery energy:
- Nominal: 12 V x 100 Ah = 1200 Wh
- Usable to 50% depth of discharge: 600 Wh
455 Wh is less than 600 Wh, so on paper the trip works, with roughly 145 Wh (about 24%) spare. That margin is thin. Lead-acid capacity falls in cold weather and falls further when current is drawn quickly, so the real usable figure is often below the label. The fridge duty cycle may be far higher than 50% on a hot day or if the van is opened often, which is the single biggest uncertainty, and the inverter on standby adds a small continuous drain. The estimate also assumes the battery starts full. Verdict: feasible but marginal, so the family should plan to drive or add a solar panel, and should treat the 50% limit as a hard floor rather than a target.
3. Evaluate the plan to run a 300 W kettle from a 300 W inverter plugged into this circuit. Include a current calculation and recommend what the family should do.
Read the worked answer
Current drawn at the battery side:
- Ideal: I = P / V = 300 / 12 = 25 A
- Allowing 85% inverter efficiency: P = 300 / 0.85 = 353 W, so I = 353 / 12 = about 29 A
Both figures are far above the 15 A fuse and above the 20 A cable rating. The fuse will blow almost immediately, which is the fuse doing its job. If the fuse were replaced with a bigger one, the 20 A cable would carry roughly 29 A, overheat and become a fire risk inside the van's lining.
The plan is therefore not viable on this circuit, and 'spare capacity on the fuse' is exactly the wrong way to think about it: the fuse rating is set by the cable, not by how much the family would like to draw. Recommendations: run the inverter on its own dedicated circuit with cable rated above 30 A, taken as a short direct run from the battery through its own fuse; keep the 15 A circuit for lights, pump, fridge and charger. They should also recognise that a 3-minute kettle boil uses only about 15 Wh, so energy is not the limit, current and wiring are. A gas or 12 V element kettle avoids the problem entirely.